GATE (TF) Textile 2016 Question Paper Solution | GATE/2016/TF/36

Question 36 (Textile Engineering & Fibre Science)

If 50 bales of 5 µg/inch, 20 bales of 3.5 µg/inch and 40 bales of 3.0 µg/inch cotton fibres are mixed, the resultant µg/inch of the mixed cotton is __3.70 : 3.85__.

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Given in the question–
Fineness of 50 bales=5 $\mu$ g/inch

Fineness of 20 bales=3.5 $\mu$ g/inch

Fineness of 40 bales=3.0 $\mu$ g/inch

Let us assume bales as $\mu$ grams
By unitary rule:
1) 50 bales of 5 $\mu$ g/inch

W₁=50 $\mu$ g
1 inch length contained=5 $\mu$ g weight

i.e , 5 $\mu$ g weight have=1 inch length

1 $\mu$ g weight have=1/5

Then,

50 $\mu$ g weight have=50/5 inch length
L₁ =10 inch

2) 20 bales of 3.5 $\mu$ g/inch

W₂=20 $\mu$ g

1 inch length contained=3.5 $\mu$ g weight

i.e , 3.5 $\mu$ g weight have=1 inch length

1 $\mu$ g weight have=1/3.5

Then,

20 $\mu$ g weight have=20/3.5 inch length
L₂ =5.71 inch

3) 40 bales of 3.0 $\mu$ g/inch

W₂=40 $\mu$ g

1 inch length contained=3.0 $\mu$ g weight

i.e , 3.0 $\mu$ g weight have=1 inch length

1 $\mu$ g weight have=1/3.0

Then,

40 $\mu$ g weight have=40/3.0 inch length
L₂ =13.33 inch

Now,
Total weight(W)=W₁+W₂+W₃
=50+20+40

Total weight(W)=110 $\mu$

Total length(L)=L₁+L₂+L₃

=10+5.71+13.33

Total length(L)=29.04 inch

Hence, 29.04 inch length have=110 $\mu$ g

1 inch length have=110/29.04

=3.79

Resultant $\mu$ g/inch=3.79 (Answer)

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